A steel safe with mass 2200kg falls onto concrete. Just before hitting the concrete its speed is 40m/s , and it smashes without rebounding and ends up being 0.06m shorter than before. What is the approximate magnitude of the force exerted on the safe by the concrete? How does this compare with the gravitational force of the Earth on the safe? Explain your analysis carefully, and justify your estimates on physical grounds.

Mass of the steel safe$m=2200\mathrm{kg}$.

Speed of the steel safe before hitting the concrete$v=40\mathrm{m}/\mathrm{s}$.

The change in length of the steel safe$D/=0.6\mathrm{m}$.

A force is a push or pull on an object caused by the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them.

The change in length of the safe is$\mathrm{D}1=0.06\mathrm{m}$. Using the approximation

role="math" localid="1656668384432" $\left(\begin{array}{c}{\stackrel{\mathrm{r}}{\mathrm{v}}}_{\mathrm{avg}}\end{array}\gg \left(\begin{array}{c}{\stackrel{\mathrm{r}}{\mathrm{v}}}_{\mathrm{f}}+\end{array}{\stackrel{\mathrm{r}}{\mathrm{v}}}_{\mathrm{i}}\right)/2=\left(\mathrm{á}0,0\tilde{\mathrm{n}}+\mathrm{á}0,‐40,0\tilde{\mathrm{n}}\right)/2=\mathrm{á},‐20,0\tilde{\mathrm{r}}\mathrm{m}/\mathrm{s}\right)$.

The time interval that the safe and the concrete were in contact is :

$$\begin{gathered} \mathrm{D}t=\frac{\mathrm{D}1}{\left|{{\displaystyle \stackrel{\mathrm{I}}{\mathrm{V}}}}_{\mathrm{avg}}\right|} \\ =\frac{0.06}{20} \\ =3\text{'}{10}^{‐3}\mathrm{s} \end{gathered}$$

Thus, the value of is $3\text{'}{10}^{‐3}\mathrm{s}$.

The force exerted is given by the formula$\stackrel{r}{F}=\frac{\mathrm{D}{\displaystyle \stackrel{\mathrm{I}}{\mathrm{p}}}}{\mathrm{D}t}$

Substituting the value of $\mathrm{D}t$as$3\text{'}{10}^{‐3}\mathrm{s}$and expanding the formula,

localid="1656669677695"

Thus, the force exerted by the safe on the concrete is .

The magnitude of gravitational force is:

$$\begin{gathered} \left|\stackrel{r}{F}\right|=\sqrt{{\left(0\right)}^2{\left(‐2.93\text{'}{10}^7\right)}^2+{\left(0\right)}^2} \\ =2.93\text{'}{10}^7\mathrm{N} \end{gathered}$$

The gravitational force on the safe is found using the formula${\mathit{F}}_{\mathbf{g}}\mathbf{=}\mathit{m}\mathit{g}$.

$$\begin{gathered} F_g=\left(2200\right)\left(9.8\right) \\ =2.16\text{'}{10}^4\mathrm{N} \end{gathered}$$

Thus, the gravitational force of the Earth on the safe is $2.16\text{'}{10}^4\mathrm{N}$.

Take the ratio of force exerted by the safe and thegravitational force of the Earth:

$$\begin{gathered} \frac{F_{impact}}{F_g}=\frac{2.93\text{'}{10}^7\mathrm{N}}{2.2\text{'}{10}^4\mathrm{N}} \\ =1.33\text{'}{10}^3 \end{gathered}$$

The force by the impact on the steel is $1.33\text{'}{10}^3$greater than the gravitational force on the safe, one factor that made the force is the time duration.